Question

Snow avalanches can be a real problem for travelers in the western United States and Canada. A very common type of avalanche is called the slab avalanche. These have been studied extensively by David McClung, a professor of civil engineering at the University of British Columbia. Suppose slab avalanches studied in a region of Canada had an average thickness of μ = 68 cm. The ski patrol at Vail, Colorado, is studying slab avalanches in its region. A random sample of avalanches in spring gave the following thicknesses (in cm).

59	51	76	38	65	54	49	62
68	55	64	67	63	74	65	79

(i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.)

x	=	cm
s	=	cm

(ii) Assume the slab thickness has an approximately normal distribution. Use a 1% level of significance to test the claim that the mean slab thickness in the Vail region is different from that in the region of Canada.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: μ < 68; H₁: μ = 68 H₀: μ ≠ 68; H₁: μ = 68     H₀: μ = 68; H₁: μ > 68 H₀: μ = 68; H₁: μ < 68 H₀: μ = 68; H₁: μ ≠ 68

(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

The standard normal, since we assume that x has a normal distribution and σ is known. The standard normal, since we assume that x has a normal distribution and σ is unknown.     The Student's t, since we assume that x has a normal distribution and σ is unknown. The Student's t, since we assume that x has a normal distribution and σ is known.

What is the value of the sample test statistic? (Round your answer to three decimal places.)

Answer 1

I HOPE ITS HELPFUL TO YOU IF YOU HAVE ANY DOUBTS PLS COMMENTS BELOW..I WILL BE THERE TO HELP YOU ...ALL THE BEST

AS FOR GIVEN DATA..

i)	x	(x-)^2
	59	7.910156
	51	116.9102
	76	201.2852
	38	567.0352
	65	10.16016
	54	61.03516
	49	164.1602
	62	0.035156
	68	38.28516
	55	46.41016
	64	4.785156
	67	26.91016
	63	1.410156
	74	148.5352
	65	10.16016
	79	295.4102
Total	989	1700.438

ii)

a) level of significance ==0.05

b)

n = 16

= 61.81

s = 10.65

Null and alternative hypothesis is

H0 : u = 67

H1 : u ≠ 67

Here population standard deviation is not known so we use t-test statistic.

Test statistic is

c)

Degrees of freedom = n - 1 = 16 - 1 = 15 , α=0.05

P-value = 0.0702        ( using t table)

P-value > α , Fail to Reject H0

I HOPE YOU UNDERSTAND..PLS RATE THUMBS UP ITS HELPS ME ALOT..

THANK YOU...!!